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Introduction to genetic analysis
Griffiths, A., Wessler, S.R., Lewontin,R.C., Gelbart, W.M.,Suzuki, D.T.
and Miller, J.H.
Eighth Edition, W.H. Freeman and Company NY
• Part I Transmission genetic analysis
Linkage maps
http://www.erin.utoronto.ca/~w3bio/bio207/index.htm
February 8
– Chapter 1: all questions p. 24-26
– Chapter 2: all the questions p. 62-72
– Chapter 3: questions #1-12,18,19, 22, 25-27, 29, 30, 32,
40-42.
– Chapter 4: sections 4.1- 4.4 and 4.6, questions # 1-4, 613,15-22,24-43.
• Part IV The nature of heritable change
– Chapter 15: sections 15.1 and 15.3; questions #1-3,1113,19,21,22, 32, 38, 52.
Neurospora crassa
• Neurospora crassa is a
haploid organism
• In N. crassa the four
products of meiosis
remain together in groups
of four called tetrads and
reflect the order of the
genes
• Each meiocyte produces
a linear array of eight
ascospores (an octad)
1
M I pattern
Chiasmata
M II pattern
• Chiasmata are the visible
manifestations of
crossovers
• Where do chiasmata
occur?
• A crossover is the
breakage of two DNA
molecules at the same
position and their
rejoining in two reciprocal
nonparental combinations
2
Four M II patterns
M II frequency
• The frequency of
octads with an M II
pattern is proportional
to the distance
between the locus
and the centromere
• M II frequency: the
frequency of meioses
with a crossover
A
A
A
A
a
a
a
a
126
Octads
a
A a A
a
A a A
a
a A a
a
a A a
A
A a a
A
A a a
A
a A A
A
a A A
132 9 11 10
Total = 300
a
a
A
A
A
A
a
a
12
Recombinant frequency
• 42 / 300 = 0.14*100 =
• M II frequency: the
14%
frequency of meioses
with a crossover
• A crossover in any
meiosis results in only • Therefore 14 / 2 = 7%
50% recombinant
chromatids (4 out of 8)
Recombinant frequency = M II frequency ⁄ 2
3
Two crossovers
• ABCxabc
–ABc
–AbC
–aBC
–abc
• Which chromatids participated in the
crossovers?
Position of crossovers
• Two crossovers involving three chromatids
Two crossovers
• ABCxabc
–ABc
–Abc
–aBC
–abC
• Which chromatids participated in the
crossovers?
4
Position of crossovers
• Two crossovers involving four chromatids
• ABCxabc
–
–
–
–
ABc
Abc
aBC
abC
Linkage maps
• T.H. Morgan first noticed that distances
between genes related to the frequency of
crossing over
• Alfred H. Sturtevant (an undergraduate
student in T.H. Morgan’s lab) calculated
the first genetic map and suggested that
recombination frequency be used as a
measure of the distance between 2 linked
genes.
Map units
• One genetic map unit (m.u.) is the distance
between gene pairs for which one product of
meiosis is 100% recombinant. So 1 m.u. is a
recombinant frequency (RF) of 0.01 ie 1%
• RF = 0.01 = 1 m.u.
• A centiMorgan cM = 1 m.u.
• For two genes 11 m.u. apart how many
recombinants would you expect from 2839
progeny
– Recombinant frequency of 11% = x ⁄ 2839
– therefore x = 305 recombinants
5
How to make linkage maps
Three-point test cross
• Check the deviation from independent
assortment
• Note the parental input genotypes
• Calculate the recombinant frequency (RF)
for each pair of loci i.e. two at a time
• Draw the map having determined the gene
order
• Calculate the interference and coefficient
of coincidence
• You are a student in a fruit fly (Drosophila
melanogaster) lab and you want to know
whether three loci are linked
• Mutant alleles
– Vermillion eyes v
– Crossveinless cv
– Cut wing edges ct
• Wild type alleles +
Linkage map
Three-point test cross
• You cross crossveinless, cut winged flies
to vermillion eyed flies (wildtype at other
loci)
– P:
– F1 :
+/+ cv/cv ct/ct X v/v +/+ +/+
+/v cv/+ ct/+
• Then you testcross female trihybrids with
tester males:
v/+ cv/+ ct/+ ♀ X v/v cv/cv ct/ct ♂
• How many genotypes do
you expect ?
• 2n = 8
• Which are the parental
genotypes for the
trihybrid?
– + cv ct
– v++
• Calculate the
recombinant frequency
• RF=284/1148 *100=
19.6%
• < 50% implies linkage
Gametes
v++
+ cv ct
v cv +
+ + ct
v cv ct
+++
v + ct
+ cv +
580
592
45
40
89
94
3
5
1448
Type
P
P
P
R
R
R
R
R
R
6
RF for each pair of loci: v & cv
• For v and cv:
• Since the parents are
v + and + cv
• Recombinants (R) are
those that are v cv
and + +
• So the RF =
• 45+40+89+94 / 1448=
268/1448= 18.5 %
Recombinant
Gametes
v & cv
v++
580
+ cv ct
592
v cv +
45
R
+ + ct
40
R
v cv ct
89
R
+++
94
R
v + ct
3
+ cv +
5
1448 268
RF for each pair of loci: v & ct
• For v and ct
• Since the parents are
+ ct and v +
• The recombinants are
those that are v ct
and + +
• So the RF for v and ct
is
• 89+94+3+5=191 / 1448
= 13.2%
• 45+40+3+5=93 / 1448
= 6.4%
Gametes
v++
+ cv ct
v cv +
+ + ct
v cv ct
+++
v + ct
+ cv +
Recombinant
cv ct
580
592
45
R
40
R
89
94
3
R
5
R
1448
Recombinant
v ct
580
592
45
40
89
R
94
R
3
R
5
R
1448 191
Draw the map
RF for each pair of loci: cv ct
• For cv and ct:
• Since the parents are
cv ct and + +
• Recombinants (R) are
those that are cv +
and + ct
• So the RF for cv & ct
Gametes
v++
+ cv ct
v cv +
+ + ct
v cv ct
+++
v + ct
+ cv +
Recombinant
Gametes
v cv
v++
580
+ cv ct
592
v ct
v cv +
45
R
+ + ct
40
R
v cv ct
89
R
R
+++
94
R
R
cv ct
R
R
v + ct
3
R
R
+ cv +
5
R
R
13.2
6.4
1448
18.5
v
13.2 m.u.
ct
6.4 m.u.
cv
13.2 + 6.4 = 19.6 which is > 18.5 !
Calculate the recombinant frequency
RF= (45+40+89+94+3+5) /1448 =
284/1148 = 0.196*100= 19.6%
7
Cross: v/+ ct/+ cv/+ ♀ X v/v ct/ct cv/cv ♂
Double crossovers should have been counted twice
Interference
• Interference (I): a
measure of the
independence of each
crossover from the other
• Coefficient of coincidence
(c.o.c): {observed
frequency or number of
double recombinants}
divided by {expected
frequency or number of
double recombinants}
• I = 1- c.o.c
I=1-
Observed frequency or number
of double recombinants
Expected frequency or number
of double recombinants
Calculate Interference
Recombinant
Gametes
v cv
v++
580
+ cv ct
592
v ct
v cv +
45
R
+ + ct
40
R
v cv ct
89
R
R
+++
94
R
R
cv ct
R
R
Observed frequency of doubles
=8
v + ct
3
R
R
+ cv +
5
R
R
13.2
6.4
1448
18.5
Observed frequency or
number of double
I = 1- recombinants
Expected frequency or
number of double
recombinants
Expected frequency of doubles =
(0.132* 0.064)1448 = 0.0084 *
1448 = 12
c.o.c = 8 ⁄ {(0.132 *0.064) 1448 = 12}
I = 1- 8 ⁄ 12
I = 12 ⁄ 12 - 8 ⁄ 12 = 4 ⁄ 12
I =⅓
8
Mapping function
• Mapping function: a function
that relates the recombinant
frequency with the mean
number of crossovers in a
segment per meiosis
• RF= ½ (1- e –m )
• Where m = average number of
crossovers in that
chromosome region
• The frequency of crossovers
can be converted to the
“corrected” RF by dividing by
2:
•
•
•
•
So for RF= 19.6% =
0.196= ½ (1- e –m )
e –m = 1-(2 * 0.196)=0.61
m = ln(.608)=0.498
• Corrected RF = m ⁄ 2
• = 0.498 ⁄ 2 = 0.249 map units
– Corrected RF= m ⁄ 2
9
Tetrad analysis
• So for fungi count the number of asci that are T type
and the number of asci that are NPD type and use
– RF= ½ T + NPD
• E.g. If you counted 100 T type asci and 2 NPD type asci
then RF= ?
• RF= ½ (100) + 2 = 5.2% or 5.2 m.u.
Parental ditype
(PD)
A· B
A· B
a· b
a· b
Tetratype (T)
A· B
A· b
a· B
a· b
Non parental
ditype (NPD)
A· b
A· b
a· B
a· B
Tetrad analysis
• To correct for double crossovers that resemble
the parents in tetrad analysis use the Perkins
formula:
– Map distance= RF=(100(NPD + ½ T)
– Corrected map distance = 50(T + 6 NPD)
Parental ditype
(PD)
A· B
A· B
a· b
a· b
Tetratype (T)
A· B
A· b
a· B
a· b
Non parental
ditype (NPD)
A· b
A· b
a· B
a· B
10
Course Overview
Outline
Week
1
2
3
4
5
6
7
8
9
10
11
12
Topic
Course objectives and Introduction to genetics
Human Pedigrees
Patterns of Inheritance: sex-linkage
Chromosomal basis of inheritance
Changes in chromosome number
Gene Mapping
Gene to Phenotype
Modified Mendelian ratios
Model organisms and mutants
Genetics of Plant Development (Arabidopsis)
Genetics of Animal Development (Drosophila)
Behaviour Genetics/Quantitative genetics
Chapter
Ch. 1 & Ch. 2
Ch. 2
Ch. 2
Ch. 3
Ch. 15
Ch. 4
Ch. 6
Ch. 6
Ch. 6 (Ch. 16)
Ch. 18
Ch. 18
Ch. 16 + papers
11